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The second and third terms of a geometric series are 192 and 144 respectively - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

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The second and third terms of a geometric series are 192 and 144 respectively. For this series, find (a) the common ratio, (b) the first term, (c) the sum to inf... show full transcript

Worked Solution & Example Answer:The second and third terms of a geometric series are 192 and 144 respectively - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

Step 1

the common ratio

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Answer

To find the common ratio, we use the relationship between the terms of a geometric series. The second term can be expressed as:

ar=192ar = 192

And the third term as:

ar2=144ar^2 = 144

Dividing the second term by the first:

ar2ar=144192\frac{ar^2}{ar} = \frac{144}{192}

This simplifies to:

r=144192=0.75r = \frac{144}{192} = 0.75

Step 2

the first term

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Answer

We can find the first term by rearranging the equation for the second term:

ar=192ar = 192

Substituting the value of r we found:

a(0.75)=192a(0.75) = 192

Solving for a, we have:

a=1920.75=256a = \frac{192}{0.75} = 256

Step 3

the sum to infinity

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Answer

The formula for the sum to infinity of a geometric series is given by:

S=a1rS_\infty = \frac{a}{1 - r}

Where a is the first term and r is the common ratio. Substituting the values we found:

S=25610.75=2560.25=1024S_\infty = \frac{256}{1 - 0.75} = \frac{256}{0.25} = 1024

Step 4

the smallest value of n for which the sum of the first n terms of the series exceeds 1000

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Answer

The formula for the sum of the first n terms of a geometric series is:

Sn=a1rn1rS_n = a \frac{1 - r^n}{1 - r}

From previous calculations:

Sn=2561(0.75)n10.75=2561(0.75)n0.25=1024(1(0.75)n)S_n = 256 \frac{1 - (0.75)^n}{1 - 0.75} = 256 \frac{1 - (0.75)^n}{0.25} = 1024(1 - (0.75)^n)

To find the smallest n such that:

1024(1(0.75)n)>10001024(1 - (0.75)^n) > 1000

This simplifies to:

1(0.75)n>100010241 - (0.75)^n > \frac{1000}{1024}

Which leads to:

(0.75)n<0.0234375(0.75)^n < 0.0234375

Taking logarithms on both sides:

nlog(0.75)<log(0.0234375)n \log(0.75) < \log(0.0234375)

This implies:

n>log(0.0234375)log(0.75)n > \frac{\log(0.0234375)}{\log(0.75)}

Calculating:

n14n \approx 14

Thus, the smallest value of n is 14.

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